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Question

The path of a projectile is given by y=ax−bx2. What is the value of the angle θ0 and speed v0 , about a point at which the projectile is launched?

A
θ0=cos11b2+1, v0=g2b(1+a2)
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B
θ0=cos11a2+1,v0=g2b(1+a2)
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C
θ0=cos11a2+1, v0=g2a(1+b2)
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D
θ0=cos11a21, v0=g2b(1+a2)
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Solution

The correct option is B θ0=cos−11√a2+1,v0=√g2b(1+a2)Given that y=ax−bx2 ……(1) The trajectory of projectile is given by y=xtan θ(1−xR) Equation (1) can be re written as y=ax(1−bxa) ⇒y=ax⎛⎜ ⎜⎝1−xab⎞⎟ ⎟⎠ ……(2) Comparing equation (1) with equation (2) we get, tanθ=a and Range of projectile, R=ab Using, 1+tan2θ=1cos2θ , we can write that, cosθ=1√a2+1 ⇒θ0=cos−11√a2+1 and, we know that, R=u2sin2θg Rewriting the above formula we get, R=u2(2tanθ1+tan2θ)g ⇒R=ab=v20(2×a1+a2)g ⇒ab=v20g(2a1+a2)⇒v20=g2b(1+a2) ⇒v0=√g2b(1+a2) Thus, option (b) is the correct answer.

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